2025 AMC 10A Problem 7

Long division would be slow here. There is a shortcut: when you divide a polynomial by $x - 1$, the remainder is simply what you get by plugging in $x = 1$. (This is the Remainder Theorem.) The same goes for $x - 2$ and $x = 2$.

Call the polynomial $P(x) = x^3 + x^2 + ax + b$.

1. Plug in $x = 1$. The remainder must be $4$:
$$1 + 1 + a + b = 4, \qquad\text{so}\qquad a + b = 2.$$

2. Plug in $x = 2$. Here $2^3 = 8$ and $2^2 = 4$, and the remainder must be $6$:
$$8 + 4 + 2a + b = 6, \qquad\text{so}\qquad 2a + b = -6.$$

3. Solve the two equations. Both have a single $b$, so subtract the first from the second to make $b$ disappear:
$$(2a + b) - (a + b) = -6 - 2, \qquad\text{so}\qquad a = -8.$$
Then $a + b = 2$ gives $b = 2 - (-8) = 10$.

4. Find what was asked. The question wants $b - a$, not $a$ or $b$:
$$b - a = 10 - (-8) = 18.$$

1. Check. With $a = -8$ and $b = 10$ the polynomial is $x^3 + x^2 - 8x + 10$. At $x = 1$: $1 + 1 - 8 + 10 = 4$. At $x = 2$: $8 + 4 - 16 + 10 = 6$. Both remainders match.

The answer is $\boxed{\textbf{(E)}\ 18}$.

Dividing by $x - c$ leaves a remainder equal to the value at $x = c$, so each remainder fact becomes one plain equation. Since $a$ and $b$ show up without any powers, those equations are easy to solve together. Next time a problem gives remainders for divisors like $x - 3$, plug in $3$ instead of dividing.

Subtract the two facts before solving anything. Going from $x = 1$ to $x = 2$, the $x^3$ part grows from $1$ to $8$ (up $7$), the $x^2$ part grows from $1$ to $4$ (up $3$), the $ax$ part grows from $a$ to $2a$ (up $a$), and $b$ does not change. The value itself goes from $4$ to $6$, up $2$. So
$$7 + 3 + a = 2, \qquad a = -8,$$
and then $1 + 1 - 8 + b = 4$ gives $b = 10$, so $b - a = 18$.